# Conditional probability

• Oct 12th 2012, 11:45 AM
sunmalus
Conditional probability
Hello, I'm stuck on a problem and I wasn't getting any answer in the pre-university statistic topic so I'd be very grateful if someone could help:

To go to work employs take their car or the bus. If they take their car they have 1/2 chances to be late, if they take the bus only 1/4 chances to be late. If they are on time one day they will take the same mean of transportation the next day, if they are late they switch. If p is the probability that an employ goes to work on day one with his car:

a) what is the probability that he'll go to work with his car on day n?

I started by writing the probability with the conditional probability that he went to work on day n-1... but that's the best i can come up with I don't know what else I can do.

b) what is the probability that he will be late on day n.

c)what is the limit when n---> inf for a) and b).

I thought of something but it doesn't work:
$C_{n}=\{\text{arrives with car on day n}\}$
$A=\{\text{arrives on time}\}$

then if $P$ is our function of probability:
$P(C_n)=P(A|C_{n-1})+P(A^{c}|C_{n-1}^c))=\frac{1}{2} P(C_{n-1})+\frac{1}{4}P(C_{n-1}^c)= \frac{1}{2} P(C_{n-1}) + \frac{1}{4}(1-P(C_{n-1})) = \frac{1}{4} (P(C_{n-1})+1)$

then recursively you get

$\frac{P(C_{1})}{4^n} + \sum^{n}_{k=1} \frac{1}{4^k}= \frac{P(C_{1})}{4^n} + \frac{1- \frac{1}{4^{n-1}}}{ \frac{3}{4}}$

but if n---->inf then $P(C_n)$ is greater than 1... find the big mistake....

• Oct 12th 2012, 04:18 PM
chiro
Re: Conditional probability
Hey sunmalus.

Are you familiar with Markov Chains?

Markov chain - Wikipedia, the free encyclopedia
• Oct 13th 2012, 01:58 AM
sunmalus
Re: Conditional probability
No I didn't but it's exactly what I needed. Thanks a lot!!
• Oct 13th 2012, 08:34 AM
sunmalus
Re: Conditional probability
btw it works the way I did it. But I forgot to put a (1/4) if front of the last expression.....
And then I find the same thing as when using Markov chains!